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| − | Consider the initial value problem (IVP)
| + | There is an alternative way of proving (a) in the Felder Script: |
| − | $$
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| − | \begin{align}
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| − | i\frac{\partial}{\partial t} f(x,t) &=-\frac{1}{2} \frac{\partial^2}{\partial x^2} f(x,t) \\
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| − | f(x,0) &= g(x) \in S(\mathbb{R}) \\
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| − | \end{align}
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| − | $$
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| − | Show that
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| − | $$
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| − | \begin{align}
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| − | f(x,t)=(\hat g(k)e^{-i\frac{k^2}{2}t})\check (x) \\
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| − | \end{align}
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| − | $$
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| − | with \(\check{\hat\phi}=\phi\) and \(\check\phi (x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\phi(k)e^{ikx}dk\) '''''(1)''''', is a solution of (IVP).
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| − | == Solution ==
| + | https://people.math.ethz.ch/~felder/mmp/mmp2/ |
| − | Rewrite the possible solution with help of '''''(1)''''':
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| − | $$ \begin{align}
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| − | f(x,t)=\left( \hat g(k) e^{-i\frac{k^2}{2}t} \right) \check (x) &=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat{g(k)} e^{-i\frac{k^2}{2}t}e^{ikx} \, dk \\
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| − | &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk
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| − | \end{align} $$
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| − | Check if the solution satisfies the initial condition:
| + | see chapter 3 -> Satz 3.1. and chapter 2 -> Satz 2.6 for a prove of Schur's lemma. |
| − | $$ f(x,0) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-0}dk = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx} \, dk =\check{\hat g}(x) = g(x) $$
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| − | For this, we used again '''''(1)''''' and the identity \(\check{\hat\phi}=\phi\).
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| − | Check if the solution satisfies the differential equation: <br />
| + | It seems to be shorter, and thus is probably better for the exam... [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 23:16, 13 June 2015 (CEST) |
| − | Calculate de derivations:
| + | |
| − | $$ \begin{align}
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| − | \frac{\partial}{\partial t} f(x,t) &= \frac{\partial}{\partial t} \left( \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \right) \\
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| − | &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\
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| − | &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}-i\frac{k^2}{2}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\
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| − | &= -\frac{i}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk
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| − | \end{align} $$
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| − | | + | |
| − | $$ \begin{align}
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| − | \frac{\partial^2}{\partial x^2} f(x,t) &= \frac{\partial^2}{\partial x^2} \left( \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \right) \\
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| − | &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial^2}{\partial x^2}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \\
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| − | &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\left(i^2k^2 \right) \hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \\
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| − | &= -\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2 \, dk
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| − | \end{align} $$
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| − | Put the whole thing in the differential equation and we see:
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| − | $$ i\frac{\partial}{\partial t} f(x,t) = -\frac{i^2}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2 \, dk =\frac{1}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk = \left( -\frac{1}{2}\right) \cdot \left( -\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk\right) = -\frac{1}{2}\frac{\partial^2}{\partial x^2} f(x,t) $$
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| − | it is satisfied.
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| − | ''Proof of exchanging the derivation and the integral:''
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| − | We want to apply the Lebesgue dominated convergence theorem. This is a standard argument being used in some other problems.
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| − | First, define \( h(x,t,k)=\hat g(k)e^{ikx-i\frac{k^2}{2}t} \). Then, as the Fourier-transform is an automorphism on the Schwartz-space:\( \hat g(k) \in \mathcal{S} ({\mathbb{R}}) \), we get considering the following argument (from Serie 11):
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| − | $$ \left| f(x) \right| \leq \frac{C}{(1 + \vert x \vert^2)} $$
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| − | for some \( C \in \mathbb{R} \) and thus:
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| − | $$ \int_{\mathbb{R}} \left| f(x) \right| dx \leq \int_{\mathbb{R}} \left| \frac{C}{(1 + \vert x \vert^2)} \right| dx < \infty $$
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| − | that \( \hat g(k) \in L^1 \).
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| − | We then want to consider
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| − | $$ \lim_{n \rightarrow \infty} \int_{\mathbb{R}} \frac{h(x + s_n, t, k) - h(x,t,k)}{s_n} \, dk $$
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| − | with \( s_n \) an arbitrary zero-sequence and see that
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| − | $$ \left| \frac{1}{s_n} \hat g (k) e^{ikx} e^{-i\frac{k^2}{2}t} \left( e^{iks_n} -1 \right) \right| = \left| \frac{2}{s_n} \hat g (k) \sin \left( \frac{ks_n}{2} \right)\right| \leq \left| \frac{2}{s_n} \hat g (k) \frac{ks_n}{2} \right| = \left| k \cdot \hat g(k) \right| \in \mathcal{S} ({\mathbb{R}}) \subset L^1(\mathbb{R} ) $$
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| − | So \( \frac{d}{dx} \int_{\mathbb{R}} h(x,t,k) \, dk = \int_{\mathbb{R}} \frac{d}{dx} h(x,t,k) \, dk \), and since \( \frac{d}{dx} h(x,t,k) \) results in a Schwartz-function depending on \( k \) times \( e^{ikx} e^{-i\frac{k^2}{2}t} \) we can apply the Lebesgue-dominated-convergence theorem twice.
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| − | It remains to be shown for \( \frac{d}{dt} \) which is basically just repeating the argument:
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| − | $$ \lim_{n \rightarrow \infty} \int_{\mathbb{R}} \frac{h(x, t + s_n, k) - h(x,t,k)}{s_n} \, dk $$
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| − | again with \( s_n \) an arbitrary zero-sequence, and then we go on:
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| − | $$ \left| \frac{1}{s_n} \hat g (k) e^{ikx} e^{-i\frac{k^2}{2}t} \left( e^{i\frac{k^2}{2}s_n} -1 \right) \right| = \left| \frac{2}{s_n} \hat g (k) \sin \left( \frac{k^2s_n}{4} \right)\right| \leq \left| \frac{2}{s_n} \hat g (k) \frac{k^2s_n}{4} \right| = \left| \frac{k^2}{2} \cdot \hat g(k) \right| \in \mathcal{S} ({\mathbb{R}}) \subset L^1(\mathbb{R} ) $$
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| − | ''Ende der Vorstellung.''
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see chapter 3 -> Satz 3.1. and chapter 2 -> Satz 2.6 for a prove of Schur's lemma.
